Question

Use the Laplace transform to solve the given differential equation subject to the indicated initial conditions.
\(y''-7y'+6y =e^t + \delta(t-3) + \delta(t-5)\)
\(y(0) = 0\)
\(y'(0) = 0\)

Answer 1

Step 1
Here we use table of Laplace transforms and some properties of Laplace transform.
Step 2
\(y''-7y'+6y =e^t + \delta(t-3) + \delta(t-5)\)
\(y(0) = 0\)
\(y'(0) = 0\)
Apply laplace transform \(\Rightarrow L\left\{y"\right\}-7L\left\{y'\right\}+6L\left\{y\right\}=L\left\{e^t\right\}+L\left\{\delta(t-3)\right\}+L\left\{\delta(t-5)\right\}\)
\(\text{use } L\left\{\delta(t-c)\right\}=e^{-cs}\)
\(\Rightarrow (s^2Y(s)-sY(0)-y'(0))-7(sY(s)-y(0))+6y(s)=\frac{1}{(s-1)}+e^{-3s}+e^{-5s}\)
\(\Rightarrow (s^2-7s+6)Y(s)=\frac{1}{(s-1)}+e^{-3s}+e^{-5s}\)
\(\Rightarrow Y(s)=\frac{1}{(s-1)(s^2-7s+6)}+\frac{e^{-3s}}{(s^2-7s+6)}+\frac{e^{-5s}}{(s^2-7s+6)}\)
Now apply inverse laplace transform
\(\Rightarrow L^{-1}\left\{Y(s)\right\}=L^{-1}\left\{\frac{1}{(s-1)(s^2-7s+6)}\right\}+L^{-1}\left\{\frac{e^{-3s}}{(s^2-7s+6)}\right\}+L^{-1}\left\{\frac{e^{-5s}}{(s^2-7s+6)}\right\}\)
Now use paricl and use table of Laplace transforms
\(\Rightarrow y(t)=-\frac{1}{6}t \cdot e^t - \frac{1}{6} e^(t-3)\cdot \delta(t-3)-\frac{1}{6}e^{t-5}\cdot \delta(t-5)\)
here we use properties \(L^{-1}\left\{F(s-c)\right\}=e^{ct} \cdot f(t)\)
\(L^{-1}\left\{e^{-cs} \cdot F(s)\right\}=y(t) \cdot f(t)\)